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What is the smallest cardinality a topology can have which is c.c.c but not separable (in ZFC)?

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Could you clarify what you mean by "cardinality of a topology"? –  François G. Dorais Mar 27 '13 at 15:43
    
Also, just to be clear, you do not require the space to be Hausdorff or have any property other than being ccc and not separable. –  François G. Dorais Mar 27 '13 at 16:05
    
Just out of curiosity, what does "c.c.c." stand for? –  Jim Conant Mar 27 '13 at 19:33
    
Countable chain condition. This means that there are no uncountable families of pairwise disjoint open sets. –  Stefan Geschke Mar 27 '13 at 20:06
    
Jim, it stands for the "countable chain condition", which could also be known as the "countable antichain condition". A topological space is c.c.c. if every family of disjoint open sets is countable. –  Joel David Hamkins Mar 27 '13 at 20:07
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3 Answers

up vote 3 down vote accepted

This may be helpful for you:

Let $X$ be a space with $|X|= \aleph_1$, let $\tau_X= \lbrace U: X\setminus U \text{ is countable } \rbrace$. This space is CCC, but not separable.

Proof: $X$ is not separable: for any countable set $A \subset X$, clearly, $U=X\setminus A$ is open and $U \cap A=\emptyset$.

$X$ is CCC: if $X$ has uncountable disjoint open sets $\lbrace \cal U_\xi: \xi \in \aleph_1\rbrace$. Pick one open set, for example, $U_0$. Because $X \setminus U_0$ is uncountable, it is a contradiction with $U_0$ is open.

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If by "cardinality of the topology" you mean "cardinality of the underlying space" then the answer is $\aleph_1$. Let $2^{\omega_1}$ be the product of $\omega_1$ many copies of the 2-point discrete space and consider the subset $X \subset 2^{\omega_1}$ consisting of all points with at most finitely many non-zero entries. Since $X$ is dense in the separable space $2^{\omega_1}$, $X$ is ccc, but it's easy to see that $X$ is not separable. Now $X$ has cardinality $\aleph_1$.

If by "cardinality of the topology" you mean "cardinality of the set of all open sets" then it depends. If you don't care about your spaces being Hausdorff, then the answer is again $\aleph_1$: just consider the topology on $\omega_1$ consisting of all final segments and the empty set. If you want your spaces to be Hausdorff, note that every infinite Hausdorff space has at least continuum many distinct open sets (just take an infinite countable pairwise disjoint family of non-empty open sets and consider all possible unions). But actually, there is a ccc non-separable space with exactly continuum many open sets in ZFC. This is Justin Moore's L-space (an example of a regular hereditarily Lindelof non-separable space). This space is ccc because it is hereditarily Lindelof and it is a subspace of the product of $\omega_1$ many copies of the circle, and hence it has a base of cardinality $\aleph_1$. Since it's hereditarily Lindelof, every open set is the union of countably many open sets from that base. Therefore it has at most continuum many open sets. So the minimum number of open sets of a Hausdorff ccc non-separable space is continuum.

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If there is a Suslin line, then there is one of size $\aleph_1$ (and indeed, Suslin trees always have size $\aleph_1$), and so one can improve the bound from continuum to $\aleph_1$, if one is caring only about the size of the underlying Hausdorff space. –  Joel David Hamkins Mar 27 '13 at 19:59
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The forcing $\text{Add}(\omega,\aleph_1)$ to add $\aleph_1$ many Cohen reals is c.c.c. but not separable, and has size $\aleph_1$, when it is considered as a poset rather than as a complete Boolean algebra. This poset is the set of finite partial functions from $\omega_1$ to $2$, ordered by extension.

But every countable space is of course separable, so the answer is $\aleph_1$.

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I'm not sure that's exactly what Niklas meant by "cardinality of a topology." I think this example has size $2^{\aleph_1}$ if we count the number of points in the compact Hausdorff space that corresponds to this example. –  François G. Dorais Mar 27 '13 at 15:43
    
Ah, the space has size $\aleph_1$, but the topology is bigger. –  Joel David Hamkins Mar 27 '13 at 15:56
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